# An Unsolvable Riccati Equation

### Problem

### Solution 1

Assume to the contrary that $y=y(x)\,$ solves the initial value problem. We shall show that this assumption leads to a contradiction.

Then, for $x\in (0,3),\,$ $y'(x)\gt 0\,$ such that $y$ is increasing and, in particular, $y(x)\gt 0,\,$ for $x\in (0,3).\,$Consider the function $z(x)=\arctan y(x).\,$ Function $z(x)\,$ is differentiable and satisfies $0\lt z\lt \displaystyle \frac{\pi}{2}\,$ and $\displaystyle \frac{z'(x)}{\cos^2x}=\tan^2z+x,\,$ i.e., $z'(x)=\sin^2x+x\cos^2x.\,$ This means that, for $x\ge 1,\,$

$z'(x)=\sin^2x+x\cos^2x\ge z'(x)=\sin^2x+\cos^2x =1,$

implying $z(x)\ge x-1,\,$ for $x\ge 1.\,$ In particular, $z(2.9)\ge 1.9\gt \displaystyle \frac{\pi}{2}.$

A contradiction.

### Investigation

I apparently found a solution, though "jumpy":

$\displaystyle y(x)=-\frac{3 \sqrt{x} \Gamma \left(\frac{2}{3}\right) J_{\frac{2}{3}}\left(\frac{2 x^{3/2}}{3}\right)}{\Gamma \left(-\frac{1}{3}\right) J_{-\frac{1}{3}}\left(\frac{2 x^{3/2}}{3}\right)}$

which works on $[0,2)\cup (2,3]$ but is undefined at $2,\,$ or $(0,3)\\{2}\,$ and verifies $y'(x)-y^2(x)-x =0.$

Verification:

### Further remarks

Mathematica can solve this directly, if you first parametrize the initial condition.Also, infinity is not exactly at 2 (but at 1.98635):

### Solution 2

Assume to the contrary that there is a solution $y=y(x).\,$ Then $y'\gt 0\,$ on $(0,3).\,$ Thus $y\gt 0\,$ and is strictly increasing on $(0,3).$

Further, $y'\gt x,\,$ implying $\displaystyle y\gt\frac{x^2}{2},\,$ so that, for $x\in [\sqrt{2},3),\,$ $y\gt 1\,$ andm therefore, $\displaystyle \frac{1}{y}\lt 1\,$ on $[\sqrt{2},3).\,$

As $y'\gt y^2,],$ $\displaystyle \left(\frac{-1}{y}\right)'=\frac{y'}{y^2}\gt 1\,$ on $(0,3),\,$ we have

$\displaystyle \frac{1}{y(\sqrt{2})}-\frac{1}{y(\sqrt{2}+1)}=\int_{\sqrt{2}}^{\sqrt{2}+1}\left(\frac{-1}{y(t)}\right)dt \gt 1.$

It follows that $\displaystyle \frac{1}{y(\sqrt{2}+1)}-\frac{1}{y(\sqrt{2})}-1\lt 0.\,$ A contradiction.

### Solution 3

Let us assume that a solution exists. Thus, for some $x$ in $(0,3)$,

$\displaystyle y(x) =\int_{0}^{x} (y^2(t)+t) dt\geq\int_{0}^{x}tdt=\frac{x^2}{2}\overset{def}{=} z_1(x).$

$\displaystyle \begin{align} y(x) &=\int_{0}^{x} (y^2(t)+t) dt \geq \int_{0}^{x}y^2(t)dt \geq \left(\frac{t^2}{2}\right)^2 \\ &= \frac{x^5}{5\cdot 2^2} \geq \frac{x^5}{2^3 \cdot 3^1} \overset{def}{=} z_2(x) \end{align}$

Let us denote these successive bounds by $z_n(x)$. We claim that these bounds can be written as

$z_n(x)=\frac{x^{p_n}}{2^{q_n}3^{r_n}},$

where $p_n$, $q_n$, and $r_n$ are sequences in $n$. We now establish the recurrence relations for these sequences.

From observation, $p_{n+1}=2p_n + 1$ (squaring doubles the power and the integration adds $1$). Noting that $p_1=2$, the solution for this linear recurrence relation is

$p_n = 3\cdot 2^{n-1}-1.$

$\displaystyle \begin{align} \int_{0}^x z_n^2(t) dt &= \int_{0}^x\left(\frac{t^{p_n}}{2^{q_n}3^{r_n}}\right)^2dt \\ &= \frac{x^{p_{n+1}}}{(2p_n+1)\cdot 2^{2q_n}\cdot 3^{2r_n}} \\ &\geq \frac{x^{p_{n+1}}}{(2p_n+2)\cdot 2^{2q_n}\cdot 3^{2r_n}} = \frac{x^{p_{n+1}}}{(3\cdot 2^n)\cdot 2^{2q_n}\cdot 3^{2r_n}} \\ &= \frac{x^{p_{n+1}}}{2^{2q_n+n}\cdot 3^{2r_n+1}} \overset{def}{=} z_{n+1}(x) \end{align}$

Thus,

$\displaystyle \begin{align} q_{n+1}&=2q_n+n;~q_1=1 \\ r_{n+1}&=2r_n+1;~r_1=0, \end{align}$

with solutions

$\displaystyle \begin{align} q_{n}&=3\cdot 2^{n-1}-n-1 \\ r_{n}&=2^{n-1}-1. \end{align}$

Thus,

$\displaystyle \begin{align} y(x)&\geq\frac{x^{(3\cdot 2^{n-1}-1)}}{2^{(3\cdot 2^{n-1} - n -1)} \cdot 3^{(2^{n-1}-1)}} \\ &\sim \left(\frac{x}{2\cdot 3^{1/3}}\right)^{3\cdot 2^{n-1}}\cdot(\text{other non-dominant terms in $n$}). \end{align}$

This bound tends to infinity as $n$ tends to infinity for $x\gt 2\cdot 3^{1/3} \sim 2.89$. This is a contradiction to our supposition that a solution exists over the entire open interval.

### Acknowledgment

This is a problem from the 1976 Moscow State University math student contest. The Investigation is by N. N. Taleb. Further remarks are due to Jyrki Penttonen. Solution 2 is by Nicu Mihalache; Solution 3 is by Amit Itagi.

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